MCQ
If a function $\text{f}:[2,\infty)\rightarrow\ \text{B}$ defined by $f(x) = x^2 - 4x + 5$ is a bijection, then $B =$
- A$\text{R}$
- ✓$[1,\infty)$
- C$[4,\infty)$
- D$[5,\infty)$
✓
Answer
Correct option: B.
$[1,\infty)$
Since $f$ is a bijection, co$-$domain of $f =$ range of $f$
$\Rightarrow B =$ range of $f$
Given: $f(x) = x^2 - 4x + 5$
Let $f(x) = y$
$\Rightarrow y = x^2 - 4x + 5$
$\Rightarrow x^2 - 4x + (5 - y) = 0$
$\because$ Discrimant, $\text{D}=\text{b}^2-4\text{ac}\geq0,$
$(-4)^2-4\times1\times(5-\text{y})\geq0$
$\Rightarrow\ 16-20+4\text{y}\geq0$
$\Rightarrow\ 4\text{y}\geq4$
$\Rightarrow\ \text{y}\geq1$
$\Rightarrow\ \text{y}\in[1,\infty)$
$\Rightarrow$ Range of $\text{f}=[1,\infty)$
$\Rightarrow\ \text{B}=[1,\infty)$
$\Rightarrow B =$ range of $f$
Given: $f(x) = x^2 - 4x + 5$
Let $f(x) = y$
$\Rightarrow y = x^2 - 4x + 5$
$\Rightarrow x^2 - 4x + (5 - y) = 0$
$\because$ Discrimant, $\text{D}=\text{b}^2-4\text{ac}\geq0,$
$(-4)^2-4\times1\times(5-\text{y})\geq0$
$\Rightarrow\ 16-20+4\text{y}\geq0$
$\Rightarrow\ 4\text{y}\geq4$
$\Rightarrow\ \text{y}\geq1$
$\Rightarrow\ \text{y}\in[1,\infty)$
$\Rightarrow$ Range of $\text{f}=[1,\infty)$
$\Rightarrow\ \text{B}=[1,\infty)$
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